Problem: Chuck the llama is tied to the corner of a $2\text{ m}$ by $3\text{ m}$ shed on a $3\text{ m}$ leash. How much area (in square meters) does Chuck have in which to play if he can go only around the outside of the shed? [asy]
draw((0,0)--(15,0)--(15,10)--(0,10)--cycle,black+linewidth(1));
draw((15,10)--(27,19),black+linewidth(1));
dot((27,19));
label("Shed",(7.5,5));
label("CHUCK",(27,19),N);
label("2",(0,0)--(0,10),W);
label("3",(0,0)--(15,0),S);
label("3",(15,10)--(27,19),SE);
[/asy]
Solution: When Chuck has the leash extended to its full length, he can move in a $270^\circ$ arc, or $\frac{3}{4}$ of a full circle about the point where the leash is attached. (He is blocked from going further by the shed.)

[asy]
draw((0,0)--(15,0)--(15,10)--(0,10)--cycle,black+linewidth(1));
draw((15,10)--(27,19),black+linewidth(1));
dot((27,19));
label("Shed",(7.5,5));
label("2",(0,0)--(0,10),W);
label("3",(0,0)--(15,0),S);
label("3",(15,10)--(27,19),SE);
draw((0,10)..(3,19)..(6,22)..(24,22)..(27,19)..(30,10)..(27,1)..(24,-2)..(15,-5),black+linewidth(1)+dashed);
draw((15,0)--(15,-5),black+linewidth(1)+dashed);
[/asy]

The area that he can play inside this circle is $\frac{3}{4}$ of the area of a full circle of radius $3,$ or $$\frac{3}{4}\times \pi(3^2)=\frac{27}{4}\pi.$$ When the leash is extended fully to the left, Chuck just reaches the top left corner of the shed, so can go no further. When the leash is extended fully to the bottom, Chuck's leash extends $1\text{ m}$ below the length of the shed. This means that Chuck can play in more area to the left.

[asy]
draw((0,0)--(15,0)--(15,10)--(0,10)--cycle,black+linewidth(1));
draw((15,10)--(27,19),black+linewidth(1));
dot((27,19));
label("Shed",(7.5,5));
label("2",(0,0)--(0,10),W);
label("3",(15,10)--(27,19),SE);
draw((0,10)..(3,19)..(6,22)..(24,22)..(27,19)..(30,10)..(27,1)..(24,-2)..(15,-5),black+linewidth(1)+dashed);
draw((15,0)--(15,-5),black+linewidth(1)+dashed);
draw((15,-5)..(11.4645,-3.5355)..(10,0),black+linewidth(1)+dashed);
label("1",(15,0)--(15,-5),W);
label("2",(15,0)--(15,10),E);
label("3",(0,10)--(15,10),N);
[/asy]

This area is a $90^\circ$ sector of a circle of radius $1,$ or $\frac{1}{4}$ of this circle.  So this additional area is $$\frac{1}{4} \times \pi (1^2)=\frac{1}{4}\pi.$$ So the total area that Chuck has in which to play is $$\frac{27}{4}\pi + \frac{1}{4}\pi = \frac{28}{4}\pi = \boxed{7\pi}\text{ m}^2.$$